# You have a "500-mL" "HCl" solution having "pH" = 3. Determine the amount of "NaOH" solid ("M"_M = "40 g/mol") that must be added so that the "pH" of the solution is changed to 10 ?

## Volume changes due to the addition of $\text{NaOH}$ are not considered.

Feb 11, 2018

$\text{0.02 g NaOH}$

#### Explanation:

Notice that the solution goes from being acidic at $\text{pH} = 3$ to being basic at $\text{pH} = 10$, so the first thing that comes to mind here is that you need to add more sodium hydroxide than the number of moles of hydrochloric acid present in the initial solution.

As you know, the concentration of hydronium cations is equal to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)}}}$

This means that the initial solution contains

["H"_3"O"^(+)] = 10^(-3) quad "M"

Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.

500 color(red)(cancel(color(black)("mL solution"))) * (10^(-3) quad "moles H"_3"O"^(+))/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.00050 moles H"_3"O"^(+)

Sodium hydroxide and hydrochloric acid react in a $1 : 1$ mole ratio to produce aqueous sodium chloride and water

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

so you know that in order to completely neutralize the acid and make the solution neutral, you need to add $0.0005$ moles of sodium hydroxide.

So you can say that at ${25}^{\circ} \text{C}$, you will have

$\text{0.0005 moles NaOH added " -> " pH} = 7$

Now, you want the target solution to have

$\text{pH} = 10$

which implies that it must have

$\text{pOH} = 14 - 10 = 4$

This means that the concentration of hydroxide anions in the final solution must be equal to

["OH"^(-)] = 10^(-4) quad "M"

Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.

500 color(red)(cancel(color(black)("mL solution"))) * (10^(-4) quad "moles OH"^(-))/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.000050 moles OH"^(-)

Therefore, you can say that if you add $0.00050$ moles of sodium hydroxide to the initial solution, you will completely neutralize the acid and get the $\text{pH}$ to $7$.

At that point, adding an additional $0.000050$ moles of sodium hydroxide will increase the $\text{pH}$ of the solution from $7$ to $10$.

The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a $1 : 1$ mole ratio--will thus be equal to

$\text{0.00050 moles + 0.000050 moles = 0.00055 moles NaOH}$

Finally, to convert the number of moles to grams, use the molar mass of the salt.

$0.00055 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaOH"))) * "40 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.02 g}}}}$

The answer is rounded to one significant figure.